What is the derivative of an iterated product like $ \frac{d}{dx}\prod \limits_{i=0}^n \ln(y_i^{x - 1})$? Suppose the following function with pi notation, with the pi denoting the iterated product, multiplying from $i = 0$ to $i = n$:
$$\prod_{i=0}^n \ln(y_i^{x - 1})$$
That is, the natural logarithm of $y$, subscripted by $i$, to the power of $x - 1$.
What is the derivative of this product - to be clear, its derivative with respect to $x$, not $y$? 
 A: We don't even need a product rule. We have
\begin{align*}
g(x) = \prod_{i=0}^{n}\ln(y_i^{x-1}) = (x-1)^{n+1}\prod_{i=1}^{n}\ln(y_i)
\end{align*}
And so
\begin{align*}
\frac{d}{dx} g(x) = (n+1)(x-1)^{n}\prod_{i=0}^{n}\ln(y_i)
\end{align*}
A: Define $\alpha=\prod_{i=0}^n \ln(y_i)$. Note that $\alpha$ is simply a constant. Use logarithm properties, use the power rule for derivatives and you're done.
$$f(x)=\prod_{i=0}^{n}\ln{y_i}^{x-1}=\prod_{i=0}^{n}(x-1)\ln{y_i}$$$$\alpha(x-1)^{n+1}\implies \dfrac{\mathrm df}{\mathrm dx}=\alpha(n+1)(x-1)^n$$
$$\boxed{\dfrac{\mathrm d}{\mathrm dx}\prod_{k=0}^n \ln{y_i}^{x-1}=(n+1)(x-1)^n\prod_{k=0}^n\ln y_i}$$
A: Let 
$$f(x)=\prod_{i=0}^{n}f_i(x)$$
and let $$g^{(m)}(x)=\left(\frac{d}{dx}\right)^mg(x),\qquad m=0,1,2,...$$
as well as $\delta_{ij}$ denote the Kronecker Delta.
We have that
$$f'(x)=\sum_{i=0}^{n}\prod_{j=0}^{n}f_j^{(\delta_{ij})}(x)=\sum_{i=0}^{n}\frac{f_i'(x)}{f_i(x)}f(x)$$
We use this with the choice 
$$f_i(x)=\ln(y_i^{x-1})=(x-1)\ln y_i$$
which gives $$f_i'(x)=\ln y_i$$
So 
$$f'(x)=\sum_{i=0}^{n}\frac{1}{x-1}\prod_{j=0}^{n}(x-1)\ln y_j$$
$$f'(x)=(x-1)^n\sum_{i=0}^{n}\prod_{j=0}^{n}\ln y_j$$
$$f'(x)=(x-1)^n\left(\sum_{i=0}^{n}1\right)\prod_{j=0}^{n}\ln y_j$$
$$f'(x)=(n+1)(x-1)^n\prod_{j=0}^{n}\ln y_j$$
